# Hall Effect Latch Help

Discussion in 'General Electronics Discussion' started by Shadowblade, Oct 7, 2014.

1. ### Shadowblade

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Oct 7, 2014
Hello! I have designed an incredibly simple circuit to demonstrate how hall effect latches work for my class. But it isn't working

I am using AH3761 hall effect latch switches (http://www.diodes.com/datasheets/AH3761.pdf) along with a cr2032 3.3v battery, a red LED and its accompanying resistor.
(red led 5mm diffused, forward current 25ma, forward voltage 1.95v).
My math is correct (3.3v - 1.95v / .25ma = 52ohm resistor.

the led with resistor lights off of the battery, but hooking up the hall sensor (led with resistor on positive lead, + lead going to OUT lead on hall switch, ground going to ground, and the battery is hooked up to the Vdd input lead on the switch and the ground on the hall switch,) and it won't work.

The switch won't work with any magnets! Am i missing something? Please help! Thank you!!!

2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,482
2,830
Jan 21, 2010
Show us a circuit diagram of what you have constructed.

From a brief look, it appears the device is open collector (or open drain) so any load needs to be connected between the output and Vcc, not between the output and ground. (page 3 illustrates this)

This will reverse the sense of the output, so the magnet will probably now turn the LED off.

You *could* connect the resistor between VDD and the output and then connect the LED between the output and ground, but the current consumption would rise (from 26mA to about 60mA) as the LED is turned off due to the device being wired to short the LED out. I would recommend that if you do it this way that you increase the value of the resistor to 100 ohms at a minimum.

A better solution would use a transistor to turn the LED on and off. But that's not quite so simple.

3. ### Shadowblade

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Oct 7, 2014
alright, this is what is hooked up at the moment (terrible paint jpeg i know, sorry. but it gets the idea across.)

File size:
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89
4. ### KrisBlueNZSadly passed away in 2015

8,393
1,271
Nov 28, 2011
Right. As Steve explained, the sensor has an open collector output and cannot drive its output positive. It can only complete the circuit between the output pin and 0V (negative supply pin).

So you have to connect the LED between the output pin and the positive supply. Take that circuit, reverse the polarity of the LED and connect the right hand wire of the LED to the end pin of the sensor (the pin that connects to battrey positive).

5. ### Shadowblade

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Oct 7, 2014
ahh i understand. This can only close on the negative side, not the positive. thats a whole aspect to electronic components i didn't even know existed.... good to know! Thanks guys. I'll be able to give it a shot when i get into the shop later. Thank you again!

6. ### Shadowblade

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Oct 7, 2014
Yepp, now they all totally work. Thank you!!!

7. ### Shadowblade

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Oct 7, 2014
Anyone interested in helping me turn this into a gerber file please? I d/l eaglecad, and even have stuff laid out and connected, but can't figure out how to export it properly. Its really only 3 parts.... you thinkd itd be easier.

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